The area of the sheet metal is 800 m^2. The length is twice as long as the the width. Let the length be L. If the width is W, L = 2W.

800 = 2*W*W

=> W = sqrt(400)

=> W = 20

L = 40

Four squares are cut...

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from the corners of the sheet of 40 x 20 cm and a box is constructed without a lid. Let the squares that are cut have a side of length s.

The volume of the box is : V = L*W*H, where L, W and H are the length, width and height of the the box.

The Length is (40 - 2s)

Width = 20 - 2s

HeightÂ = s

Volume = (40 - 2s)(20 - 2s)(s)

=> (40s - 2s^2)(20 - 2s)

=> 800s - 40s^2 - 80s^2 + 4s^3

=> 4s^3 - 120s^2 + 800s

To maximize volume find the derivative and solve for s.

12s^2 - 240s + 800 = 0

=> 3s^2 - 60s + 200 = 0

s1 = 15.77

s2 = 4.22

There are two roots of the equation, one giving a minimum point and the other the maximum point. The second derivative of the volume which is 24s - 240 is used to determine which of the values of s give the maximum. 24s - 240 has a negative value for the value of s that gives a maximum value.

For s = 15.7, 24s - 240 = 138.48

and for s = 4.2 , 24s - 240 = -138.72

This indicates that volume is minimum if s = 15.77 and maximum if s = 4.2.

For s = 4.2, the dimensions of the box are: 31.6, 11.6 and 4.2

**The box has a maximum volume when the dimensions are 31.6, 11.6 and
4.2 cm**